$N_{2} + 3 H_{2} ⇌ 2 {NH}_{3}$ , $K = 4 \times 10^{8} at 298 K$ , $K = 41 at 400 K$ Which statements is correct?

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If $N_{2}$ is added at equilibrium condition, the equillbrium will shift to the forward direction use according to Il nd law of thermodynamics the entropy must increase in the direction of spontaneous reaction

The condition for equilibrium is $2 Δ G_{{NH}_{3}}$ $= 3 Δ G_{H_{2}} + Δ G_{N_{2}}$ where $G$ is Gibbs free energy/ per mole of the gaseous species measured at that partial pressure

Addition of catalyst does not change $K_{p}$ but changes $Δ H .$

At 400 K addition of catalyst will increase forward reaction by 2 times while reverse reaction rate will be changed by 1. 7 times.

answer is B.

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Detailed Solution

At equilibrium:

$Δ G_{R \times n} = 0 or 2 Δ G_{{NH}_{3}} - Δ G_{N_{2}} - 3 Δ G_{H_{2}} = 0$

For this equation :

$N_{2} + 3 H_{2} ⇌ 2 {NH}_{3}$ , $K = 4 \times 10^{8} at 298 K$ , $K = 41 at 400 K$ correct statements is

The condition for equilibrium is $2 Δ G_{{NH}_{3}}$ $= 3 Δ G_{H_{2}} + Δ G_{N_{2}}$ where $G$ is Gibbs free energy/ per mole of the gaseous species measured at that partial pressure.